Question

我有两个标记对象列表：

class Marker():
    def __init__(self,marker_number, marker_data):
        self.marker_number = marker_number
        self.marker_data = marker_data

我想做的是：

>>> existing_markers = [Marker(marker_number=1, marker_data= 'a'),
                     Marker(marker_number=2, marker_data= 'b'),
                     Marker(marker_number=3, marker_data= 'c'),
                     Marker(marker_number=4, marker_data= 'd'),
                     Marker(marker_number=5, marker_data= 'e'),]

>>> new_markers = [  Marker(marker_number=1, marker_data= 'aa'),
                     Marker(marker_number=3, marker_data= 'bb'),
                     Marker(marker_number=5,  marker_data= 'cc'),]

>>> interlace_markers(existing_markers, new_markers)

[Marker(marker_number=1, marker_data= 'a'),
Marker(marker_number=2, marker_data= 'aa'),
Marker(marker_number=3, marker_data= 'b'),
Marker(marker_number=4, marker_data= 'c'),
Marker(marker_number=5, marker_data= 'bb'),
Marker(marker_number=6, marker_data= 'd'),
Marker(marker_number=7, marker_data= 'e'),
Marker(marker_number=8,  marker_data= 'cc')]

因此，当我交织标记时，new_markers列表中与原始标记列表具有相同编号的标记出现在原始标记之后，但编号更新以维护编号序列。这是我目前的做法：

def interlace_markers(current_markers_list, new_markers_list):

    interlaced_markers = []
    #interlace the markers
    for existing_marker in current_markers_list:
        interlaced_markers.append(existing_marker)
        for new_marker in new_markers_list:
            if new_marker.marker_number== existing_marker.marker_number:
                interlaced_markers.append(new_marker)

    #reset the sequence
    sequence_index = 1
    for marker in interlaced_markers:
         marker.marker_number= sequence_index
         sequence_index += 1

    return interlaced_markers

我有两个问题：

最狡猾的方式是什么
原始列表可能很大~10k，并且是从一个外部文件，但新列表将小~100-300长，是有一种更有效的方法，而不是使用列表？

Answer 1

如果我理解正确，我认为通过简单的排序最容易实现：

interlace_markers = existing_markers + new_markers
interlace_markers.sort(key = lambda x: x.marker_number)

现在，如果您想更改编号，可以在一个简单的循环中执行此操作：

for i,marker in enumerate(interlace_markers):
    marker.marker_number = i+1 #indexing is 1 based instead of zero based

请注意，这是有效的，因为python的排序是 stable ，这意味着如果没有必要，订单不会被更改。

这导致了一个O（n）+ O（nlogn）= O（nlogn）算法，这个算法并不算太差。

基于python中列表对象的属性隔行两个对象列表

1 个答案: